Direct Current Circuits

The following simple question appeared in Kerala Medical Entrance 2006 test paper:
Four cells each of emf 2V and internal resistance 1Ω are connected in parallel to a load resistor of 2Ω. Then the current through the load resistor is
(a) 2A (b) 1.5A (c) 1A (d) 0.888A (e) 0.75A
Since the cells are connected in parallel, the internal resistance of the combination is 1/4=0.25Ω. If ‘R’ is the load resistance and ‘r’ is the internal resistance, the current through load resistance= V/(R+r) = 2/(2+0.25) = 0.888A. So (d) is the correct option.
Now consider the following M.C.Q. which appeared in the question paper of Karnataka C.E.T.2003:
Two wires of the same dimensions but resistivities ρ₁ and ρ₂ are connected in series. The equivalent resistivity of the combination is
(a) 2(ρ₁ + ρ₂) (b) √(ρ₁ρ₂) (c) (ρ₁ + ρ₂)/2 (d) ρ₁ + ρ₂
If R₁ and R₂ are the resistances of the wires, R₁ = ρ₁L/A and R = ρ₂L/A where L and A are the length and the cross section area respectively. When the wires are connected in series the total length is 2L and the total resistance is R₁ + R₂. Hence the resistivity of the combination is given by ρ = (R₁ + R₂)A/2L = (R₁A/2L) + (R₂A/2L) = ρ₁/2 + ρ₂/2. The correct option therefore is (c).
Suppose we make this question a bit more complicated by making the lengths of the two wires different (say, L₁ and L₂). Then, ρ = (R₁ + R₂)A/(L₁ + L₂) = [(ρ₁L₁/A) + (ρ₂ L₂/A)] A/(L₁ + L₂) = (ρ₁L₁ + ρ₂L₂)/(L₁ + L₂).
Now, consider the following question:
The heat produced per second in a 2Ω resistor on connecting it across a battery is the same as the heat produced per second in an 8Ω resistor on connecting it across the same battery after disconnecting the 2Ω resistor. The internal resistance of the battery is
(a) 4Ω (b) 5Ω (c) 6Ω (d) √10 Ω (e) 3.3Ω
It is convenient to remember the internal resistance as √(R1R2) in situations like this. The answer is √(2×8) = 4 Ω. You can work it out from ‘first principles’ like this:
The currents through the resistors in the two cases are V/(2 + r) and V/(8 + r) with usual notations. Equating the powers (I2R values), we have, [V/(2 + r)]2×2 = [V/(8 + r)]2×8, from which r = 4 Ω.
Let us consider the following question also taken from Karnataka C.E.T.2003 question paper:
If a 30V, 90W bulb is to be worked on 120V line, the resistance to be connected in series with the bulb is
(a) 40 Ω (b) 30 Ω (c) 20 Ω (d) 10 Ω
30V, 90W bulb means the power consumed by the bulb on connecting it across a 30volt supply is 90 watt. The current drawn by this bulb (while incandescent) is 90watt/30 volt = 3 ampere and the resistance of this bulb is 30V/3A = 10 Ω. The problem is basically to find out the resistance required to limit the current to 3A. If R is the resistance required in series with the bulb, we have, 120volt/(10 + R)ohm = 3 ampere. Therefore, R = 30 Ω.

All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2007- Dates Announced

Central Board of Secondary Education (CBSE), Delhi has announced the dates pertaining to the All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2007 for admission to 15% of the total seats for Medical/Dental Courses in all Medical/Dental colleges run by the Union of India, State Governments and Municipal or other local authorities in India except in the States of ANDHRA PRADESH AND JAMMU & KASHMIR. The dates of the Examination are:
(1) Preliminary Examination : 1st April 2007 (Sunday)
(2) Final Examination : 13th May 2007 (Sunday)

The Information Bulletin and Application Form costing Rs.400/- for General Category candidates and Rs.200/- for SC/ST candidates inclusive of examination fee can be obtained against Cash Payment from designated branches of Canara Bank and Regional Offices of CBSE up to 18-11-2006. Designated branches of Canara Bank in Kerala are:
KOTTAYAM : P.B. No.122, K.K.Road, Kottayam-686 001

QUILANDY : Fasila Buidling, Main Road, Quilandy-673 305

TRIVANDRUM : Ist Floor, Ibrahim Co. Bldg., Challai, Trivandrum-695 023
TRIVANDRUM : Plot no.2, PTP Nagar Trivandrum-695 038
TRIVANDRUM : TC No.25/1647, Devaswom Board Bldg. M.G. Road, Trivandrum-695 001

ERNAKULAM : Shenoy’s Chamber, Shanmugam Road, Ernakulam, Cochin-682 031
CALICUT : 9/367-A, Cherooty Road, Calicut-673 001
TRICHUR : Trichur Main Ramaray Building, Round South, Trichur-680001
QUILON : Maheshwari Mansion, Tamarakulam, Quilon-691 001
PALGHAT : Market Road, Big Bazar, 20/68, Ist floor,Palghat-678 014.

The Information Bulletin and Application Form can also be obtained by Speed Post/Registered Post by sending a written request with a Bank Draft/Demand Draft for Rs.450/- for General Category and Rs.250/- for SC/ST Category payable to the Secretary, Central Board of Secondary Education, Delhi along with a Self Addressed Envelope of size 12” x 10”. The request must reach the Deputy Secretary (AIPMT), CBSE, 2, Community Centre, Preet Vihar, Delhi-110 092 on or before 10-11-2006. The request should be super scribed as “Request for Information Bulletin and Application Form for AIPMT, 2007”.

Completed Application Form is to be dispatched by Registered Post/Speed Post only. Last Date for receipt of completed Application Forms in CBSE is 20-11-2006. You can obtain complete information at www.cbse.nic.in

Rotational Motion – Rolling bodies

If you release from rest differently shaped regular bodies such as disk, ring, hollow sphere, solid sphere, hollow cylinder(pipe) and solid cylinder from the top of an inclined plane, thereby allowing them to roll down the plane, you will find that the solid sphere always arrives at the bottom first and the ring and the pipe arrive last. The acceleration ‘a’ of a body of radius ‘R’ rolling down an inclined plane (without slipping) is given by a= g sinθ/[1+( k² / R²)] where ‘g’ is the acceleration due to gravity, ‘θ’ is the angle of the plane and ‘k’ is the radius of gyration given by I= Mk² where ‘I’ is the moment of inertia of the body (of mass M) about the central axis about which it is rolling down. Since the moment of inertia of a solid sphere about its diameter is (2/5) MR², we have (2/5)MR² = Mk². Therefore, k²/R² = 2/5.
The ring and the pipe have k²/R² = 1, since I=MR² = Mk². If you find the values of k²/R² in the case of differently shaped rolling bodies, you will realize that it is minimum (2/5) for the solid sphere and maximum (equal to1) for a thin ring and a thin pipe. Since (k²/R²) appears in the denominator in the expression for ‘a’, the acceleration down the plane is maximum in the case of a solid sphere and minimum in the case of a ring or a pipe. So the solid sphere arrives at the bottom first and the ring and the pipe arrive last.
You should remember that the moment of inertia of a disk (and that of a solid cylinder) about its rolling axis is ½ MR² and that of a hollow sphere is (2/3) MR²
It is interesting to note that all solid spheres will arrive at the bottom together, irrespective of their mass and size. Similarly, all thin rings and thin pipes will arrive together, irrespective of their mass and size. Generally speaking, bodies of a given shape will arrive together, irrespective of their mass and size.
Now, consider the following M.C.Q.:

A body of mass M at the top of a smooth inclined plane starts from rest and slides down the plane. It reaches the bottom with a velocity ‘v’. If the same body were in the form of a ring and the plane were rough, the velocity with which the ring will reach the bottom on rolling down the plane would be
(a) v (b) v√2 (c) 2v (d) v/√2 (e) v/2
Note that if a body is to roll along any surface, friction is necessary and that is why the plane is said to be made rough in the problem. On a smooth plane, the body will slide down with acceleration gsinθ and will reach the bottom with a velocity ‘v’ given by v² = 0 + 2gsinθ×s, as given by the usual equation of linear motion, v² = u² + 2as.
Therefore, v = √(2sgsinθ).
On rolling, the acceleration down the plane is gsinθ/[1 + (k²/R²)] = ½ gsinθ since k²/R² = 1 for a ring. Therefore the velocity (v1) on reaching the bottom is given by v₁² = 0 + 2×½ gsinθ×s
Therefore, v₁ = √(sgsinθ) = v/√2 [Option (d)].

Let us discuss another M.C.Q.:

A sphere of radius ‘R’ is kept at the top end of a curved track. The upper end of the track is at a height ‘H’ and the lower end which is horizontal, is at a height ‘h’ above above the ground level.

When the sphere is released, it rolls down (without slipping) along the track and after leaving the lower end of the track, it moves like a projectile and lands at the point C. The horizontal distance BC is
(a) R√[(20/7)gh(H-h)] (b) √[(20/7)gh(H-h)] (c) √[(10/7)gh(H-h)]
(d) √[(20/7)h(H-h)] (e) √[(5/7)h(H-h)]
The loss of potential energy by the sphere on rolling down along the track is equal to the gain of kinetic energy (both translational and rotational) so that
Mg(H-h) = ½ Mv² + ½ I ω² where M is the mass, ‘v’ is the linear velocity, ‘ω’ is the angular velocity and ‘I’ is the moment of inertia of the sphere, which is (2/5)MR². Substituting for ω = v/R, we obtain v = √[(10/7)g(H-h)]. This is the horizontal velocity of the sphere at the lower end of the track. Since its height at the lower end of the track is ‘h’, the time taken to reach the ground from there is t = √(2h/g). Note that we obtain this time by considering the vertical motion of the sphere and using the equation, h = 0 + ½ gt². The horizontal distance BC = vt = √[(20/7)h(H-h)]. So, (d) is the correct option. Note that this contains neither R nor g.
The following question is a conventional type which you will get in your class examinations as well as in entrance tests:

A solid cylinder of mass M and radius R rolls down an inclined plane of height ‘h’ without slipping after starting from rest at the top. The speed of its centre of mass when it reaches the bottom is
(a) √(2gh) (b) √(4gh/3) (c) √(3gh/4) (d) √(2gh/3) (e) √(3gh/2)
Equating the gravitational potential energy of the cylinder at the top of the plane to the sum of the translational and rotational kinetic energies at the bottom, we have,Mgh = ½ Mv² + ½ I ω² = ½ Mv² + ½ ×½ MR² × (v/R)² = (3/4)Mv², from which v = √(4gh/3).

Here is another typical M.C.Q. involving rotational motion:
A spherical ball rolls on a table without slipping. The fraction of its total energy which is associated with rotational motion is
(a) 3/5 9b) 2/3 (c) 2/5 (d) 3/7 (e) 2/7
The rolling ball has translational and rotational kinetic energies giving it total kinetic energy equal to ½ Mv² + ½ I ω² = ½ Mv² + ½ ×(2/5)MR² × (v/R))² =½ Mv² + (1/5)Mv² = (7/10)Mv².
The rotational kinetic energy is (1/5)Mv² and the total kinetic energy is (7/10)Mv². The required ratio is therefore (1/5) / (7/10) which is 2/7 [Option (e)].

You will see interesting questions (with solution) on rotational motion here as well as here.

If you would like to have still more multiple choice questions (with solution) on rotational motion, you can have them here.

Kinetic energy of gas molecules

Most of you might be remembering the expression for the total average kinetic energy of a gas molecule: E = (n/2)kT where ‘n’ is the number of degrees of freedom, ‘k’ is Boltzman’s constant and T is the absolute temperature. The values for ‘n’ are 3 for mono atomic, 5 for diatomic and 6 for tri and polyatomic gas molecules. [Note that we have not considered the possible vibrational modes of diatomic and polyatomic gas molecules here].
One important thing you have to note is that the average translational kinetic energy of all types of molecules is (3/2)kT since we have a universe with three dimensional space (n=3).
If you consider one mole of gas molecules, the average total kinetic energy (for one mole) is (n/2)RT and the average translational kinetic energy (for one mole) is (3/2)RT where ‘R’ is universal gas constant.
Now, consider the following M.C.Q.:
The average translational kinetic energy of oxygen molecule at a temperature ‘T’ is 6×10^-21 J. The average translational kinetic energy of helium atom at a temperature 2T will be
(a) 3.6×10^-21 J (b) 7.2×10^-21 J (c) 2.88×10^-20 J (d) 6×10^-21 J (e) 1.2×10^-20 J
Since the translational kinetic energy is (3/2)kT for all types of gas molecules, it depends only on temperature and the answer is 1.2×10^-20 J [Option (e)].
The following MCQ high lights the type of the molecule in addition to the temperature you will have to consider while calculating the kinetic energy:
The ratio of the total kinetic energy of all the molecules in one mole of hydrogen at temperature ‘T’ to the total kinetic energy of all the molecules in two moles of helium at temperature 2T is
(a) 5:6 (b) 5:12 (c) 5:3 (d)1:2 (e) 1:4
The total kinetic energies per mole in the case of hydrogen and helium are, respectively, (5/2)RT and (3/2)RT since hydrogen is diatomic and helium is mono atomic. Hence the required ratio is [(5/2)RT] : [2×(3/2)R×2T] = 5:12 [Option (b)].
The following question appeared in the I.I.T.1997 test paper:
The average translational energy and the rms speed of molecules in a sample of oxygen gas at 300 K are 6.21×10^-21J and 484 m/s respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour)
(a) 12.42×10^-21J, 968 m/s (b) 8.78×10^-21J, 684 m/s (c) 6.21×10^-21J, 968 m/s (d) 12.42×10^-21J, 684 m/s
Average translational energy per mole = (3/2)RT . When the temperature changes from 300 K to 600 K, the energy is doubled. The rms speed of gas molecules = √(3kT/m) where ‘k’ is Boltzman’s constant and ‘m’ is the molecular mass. So rms speed becomes √2 times when the temperature changes from 300 K to 600 K. The correct option is (d).

You can find more multiple choice questions at physicsplus: Questions on Kinetic Theory of Gases